Can. J. Math., Vol. XXXVI, No. 2, 1984, pp. 344-360
ASSOCIATED PRIME IDEALS IN NON-NOETHERIAN RINGS
JUANA IROZ AND DAVID E. RUSH
The theory of associated prime ideals is one of the most basic notions in the study of modules over commutative Noetherian rings. For modules over non-Noetherian rings however, the classical associated primes are not so useful and in fact do not exist for some modules M. In [4] [22] a prime ideal P of a ring R is said to be attached to an JR-module M if for each finite subset / of P there exists m e M such that / Q ann# (m) Q P. In [4] the attached primes were compared to the associated primes and the results of [4], [22], [23], [24] show that the attached primes are a useful alternative in non-Noetherian rings to associated primes. Several other methods of associating a set sé(M) of prime ideals to a module M over a non-Noetherian ring have proven very useful in the past. The most common of these is the set Assy(M) of weak Bourbaki primes of M [2, pp. 289-290]. Another method, which was used by Krull in 1929 [8] and later studied by Banaschewski [1] and Kuntz [9], is the following. Call a prime ideal Pa Krull prime of M if for each a e P there exists an m e M such that a e ann# (m) Q P. In his study of pseudo-Noetherian rings [14] [15], K. McDowell was led in [16, pp. 36-37] to consider the set of attached primes of M, which he called the strong Krull primes of M due to their relationship to the Krull primes. We prefer McDowell's terminology because of this connection with the Krull primes and also because the "attached prime" terminology has previously been used to mean something entirely different. (See for example the papers [12, 13, 25, 27, 29, 30, 31].) McDowell developed many of the basic properties of strong Krull primes in [16] and in [17] where he compared the strong Krull primes to other well-known types of associated primes that have been used in non-Noetherian rings. After defining and summarizing in Section 1 some of the relationships among seven different notions of associated primes which appear in the literature, we show, in Section 2, that unlike the Nagata primes and weak Bourbaki primes, the Krull primes and strong Krull primes behave well with respect to flat ring extensions. Some further results on the behavior of these primes in polynomial rings are also given. After some remarks in Section 3 on when the strong Krull primes and weak Bourbaki primes are the same, we give in Section 4 a consequence of this behavior for
Received November 25, 1982 and in revised form August 5, 1983.
344
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seminormality in abelian group rings. In Section 5 we show that the main result of [23] can be strengthened by replacing the strong Krull primes by weak Bourbaki primes. In the final section we give a brief discussion of the notion of "associated prime" in general. The authors were fortunate to have had access to the unpublished manuscript [17] and gratefully acknowledge the debt that the present work owes to it.
1. Definitions and notation. We begin by giving the definitions of seven different notions of associated prime ideals that appear in the literature. For this we will need some notation. All rings will be commutative with identity. If M is an ^-module we let
ZR(M) = {r e R\rm = 0 for some m e M, m ¥= 0}.
If S is a multiplicative subset of R, M (S) denotes
{m If S = R — P where P is a prime ideal of R we write M(P) instead of M(S). UN ç M, ami;? (N) denotes the annihilator of N(= {x <= R\xN = 0} ). We will write ann# (m) instead of ann# ( {m} ) if m e M. Definition. Let P be a prime ideal of R, and M an i^-module. (a) If P = ann# (m) for some m e M we call P a Bourbaki prime of M. We let Ass(M) denote the set of Bourbaki primes of M. See for instance [2, Chapter IV] for a discussion of these primes. (b) If P Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 346 J. IROZ AND D. E. RUSH the basic properties of these primes can be found in [20] [33]. We will denote the set of Nagata primes of M by N(M). For an i^-module M we have Ass(M) ç Z-S(M) ç Ass/(M) Q sK(M) Ç K(M) Q N(M) and Ass(M) Q Ne(M) ç Assy(M). Further, all of these containments can be proper [33] [9] [10] [17], and there are no inclusions relating Ne(M) and Z — S(M). The equality Z-S(M) = N(M) holds if the zero submodule of M has a primary decomposition [20], and if R is Noetherian then Ass(M) = N(M). Also, it can happen that Ass(M), Ne(M), and Z — S(M) are empty for M ^ 0, whereas clearly Ass/(M) = 0 ** M = 0. In [9] the Krull primes of M were studied only for cyclic modules. However, most of these results extend easily to arbitrary i?-modules M. For example, if M is an i?-module and P is a prime ideal, the following statements are equivalent (see [9, Proposition 1] ): (a) P G K(M\ (b)P ç ZR{M/M(P)\ (c)P = ZR(M/M(P)). Also, if S is a multiplicative subset of R it follows that l l Ks-iR(S~ M) = {S~ P\P e KR(M), P n S = 2. Associated primes and ring extensions. In this section Ass/^M) ç ^(AsS/(M) ) and that equality holds if Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 347 a Ass,(*(M 0^ B) = fl P G #Ass/(M ®A B) ) with P £ Assy(M), i.e., weak Bourbaki primes of M ®A B do not necessarily contract to weak Bourbaki primes of M. More recently Heitmann has given an example showing that Nagata primes of M ®A B do not always contract to Nagata primes of M [6] (again where (j>:A —> B is flat). To date the only positive result obtained without further restrictions on (2.1) PROPOSITION. If M is a B-module, then sK(^M) = ^(sK(M) ). Proof. Let Q e sK(M), P = a4>(Q) and let I be a finitely generated ideal of A contained in P. Then IB Q PB Q Q so there exists m e M such that IB ç annfi (m) Q Q. Thus _1 / c ^ (ann5 (m) ) = ann^ (m) Q P and P G SK(^M). For the opposite inclusion let P G SK(^M). Since strong Krull primes localize well, we can assume that A is quasi-local with maximal ideal P. Then PB ¥= B, for otherwise we would have n 1 = 2 afti with cij G P and bt G B. i=\ But then since P G SK(^M) we would have m G M such that [ah . . ., an) Q ann^ (m) Q P and n m = \m = Zu aftim = 0 z = l which is impossible. Let Q G Spec(2?) be minimal over PB. By localizing at 0 we may assume that (B, Q) is quasi-local and Q = radical of PB. Then if / #= 0 is a finitely generated ideal of B contained in Q there exists Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 348 J. IROZ AND D. E. RUSH an integer n ^ 1 and a finitely generated ideal J oî A contained in P with n I Q JB Q PB. Since P e SK(^M) there exists m <= M such that J Q ann^ (m) Q P. Then r Q JB Q ann^ (m)B Q ann5 (m) Q Q. Choose n such that Inm = 0 but In~xrn * 0 (where 7° = ,4). Then if x e 7"_1 with xm ^ Owe get 7 Q ann# (xm) ç g and hence Q e SK(M). Remark. The above proof actually shows that any prime which is minimal over PB is in sK(M). The next theorem was given by McDowell for the case that A is coherent and M is a finitely presented ,4-module. An application will be given in Section 4. (2.2) THEOREM. If(j>:A —> B is flat and M is an A-module, then >(sK(M®AB))Q sK(M) and equality holds if (f> is faithfully flat. a Proof Let Q e sK(M ®A B) and let P = 0 ¥= Hom5 (BUB, M ®A B) = Hom^ (y*//, M) ®A B\ the isomorphism from [2, p. 23, Proposition 11]. Thus Hom^ {All, M) ¥= 0 since B is faithfully flat. For the last statement recall that if cj> is faithfully flat, the map M -> ^(M ®^ 5) is injective [2, p. 32, Proposition 8]. Thus a sK(M) c sK(^(M 0^ B) ) = a sK(M) = Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 349 (2.3) PROPOSITION. Let (2.4) THEOREM. If§\A -> B is aflat A-algebra and M is an A-module, then a (j>(K(M ®A B) ) c K(M). a Proof. Let Q G K(M ®A B) and let P = 0 -> annM (JC) ®^ 5 -» M 0^ 5 *-> M ®A B. Thus we get annM (x) ®^ 5 = annM05 (x 0 1) = annM0# ( and hence annM (x) ¥= 0. We conclude this section with a couple of results on associated primes in polynomial rings. The first shows a further relationship between Krull primes and strong Krull primes. The second will be useful in Section 5. (2.5) THEOREM. Let M be an R-module and X an indeterminate. Then [PR[X] \P G sK(M) } = sK(M 0* R[X] ) = K(M ®R R[X] ). Proof. We first show [PR[X] \P G sK(M) } ç SK(M ® R[X] ). Let/i, . . . ,f„ G PP[X], P G sK(Af). Then there exists m G M, such that m annihilates the coefficients of the/ and (0:Rm) Q P. Let {/,,...,/„} Ç (0:*m K(M ®R R[X] ) ç [PR[X] \P G sK(M) } we first show that if Q G K(M 0 R[X] ) then g is an extended prime; g = PR[X] where P = g n P. For this we may assume R = RP. If Q ¥= Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 350 J. IROZ AND D. E. RUSH PR[X], then Q = (P,f)R[X] where/ G R[X] is monic and irreducible modulo PR[X]. But then /e Q G ÂXM®*P[X]) implies that / is a zero divisor on M[X] and this is impossible since / is monic. Thus Q is an extended prime. It remains only to show that if PR[X] G K(M ®R R[X] ) then P G sK(Af), and for this we may again assume that R = RP. Let a& a\, . . . , an G P. To show that there exists m G M such that ÛQ, . . ., an G (0:Rm) Q P let / = a0 + fllZ + . . . + anX\ Then/ G PPIT] so there exists g = m0 + mxX + . . . + mKtf G M[X] = M ®R R[X] such that /e(0W)c?liffl. By [21, Theorem 1] there exists an integer; such that +i c(fy c(g) = c(fyc(fg) = o where c(h) denotes the P-submodule generated by the coefficients of h. If j is chosen minimal such that c(fyJrXc(g) = 0, then for any non-zero x G c(fy'c(g) we have (a0,...,an) = c(f) Q (0:Rx) Q P. Thus P G sK(M). If / is an ideal of R we will let ^1 = {a ^ R\an^ I for some n â 1}. (2.6) THEOREM. Let M be an R-module and X an indeterminate. Then Assf(M ®R R[X]) = [PR[X] \P G Ass/(M) }. Proof. Let Q G ASS/(M ®R R[X] ) and P = Q n R. Since g G SK(M ® P[X] ), then Q = PR[X] by Theorem 2.5. To show that P G Assy(M) we may assume that R = RP. Assume that PR[X] = Q is minimal over (Q'R[X]g) where g = m0 + m^ + . . . + m„.T G M[X] = M ® P[X], mz G M. First we show that P Q V(®'Rc(g) )• Let ^ ^ ^- Then a G P#[X] implies there exists k G R[X] - PP[X] such that A:a" G (Oi/^g) for some n â 1. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 351 Thus a" kg = 0 and since c(k) = R we get 0 = c{ankg) = a"c(g). Thus P c ^0:Rc(g). But then P Q V(0:«c(g)) = n y/Ôw) Q P, and thus P = \/(0\Rmi) for some /*. It remains to show that if P e Assy(M), then PR[X] <= Ass/(M ®R R[X] ). Again it suffices to consider the case R = RP. Then P e Assy(M) => P = ^(O'.Rm) for some m e M. Cte'm. />£[*] = \/0:/?[ATW. Let/ G PR[X]. Since c(/) Q VW^w) we can choose n â 1 such that c{f)nm = 0. Then c(fnm) = c(fn)m Q c{f)nm = 0 and hence f ^ V(0-R[X]*n). Conversely, assume/ £ PR[X]. Then c(fn) = R for all « > 0. But then if fnm = Owe get 0 = c(fnm) = c(/w)w = m, a contradiction. Thus />*[*] = VÔ^n- 3. Assy and sK. Since for many purposes it is desirable to have P e Assy(M) rather than just P e SK(M) (e.g. as in the proof of Theorem 4.2) and since sK(M) is in some ways better behaved than Assy(M), it would be of interest to know when Assf(M) = sK(M). Lacking a useful ideal-theoretical characterization of the rings R with the property that Assf(M) = sK(M) for every i^-module M we consider the following obviously sufficient condition for this to hold: (MFG) Each prime ideal of R is minimal over a finitely generated ideal of R. Perhaps the most obvious examples of non-Noetherian rings with this property are the rings with Noetherian spectrum. Recall that this means that R has the following equivalent properties [5]: (3.1) (i) R has the ascending chain condition on radical ideals. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 352 J. IROZ AND D. E. RUSH (ii) For each ideal A there exists a finitely generated ideal A0 with V^ = VÂ~O. (iii) For each prime ideal P there exists a finitely generated ideal A such that P = yZ (iv) R has the ascending chain condition on prime ideals and every ideal of R has finitely many minimal prime divisors: We make a few simple observations on the rings satisfying MFG. (3.2) (i) A von-Neumann regular ring R satisfies MFG, but has Noetherian spectrum only if it is Noetherian. Proof Let m be a non-finitely generated maximal ideal of R. Then m is minimal over the zero ideal, but is not the radical of any finitely generated ideal since any finitely generated ideal of R is contained in a principal maximal ideal. (ii) If R satisfies MFG, then R has the ascending chain condition on prime ideals. The converse holds ifSpec(R) is a tree (i.e., if for each Q £ Spec(/?), {P G Spec(i?)|P ç Q} is a chain). Proof Let P\ Q P2 Q ... be a chain of prime ideals of R. Then Q = CO U /=! Pj is a prime ideal so Q is minimal over a finitely generated ideal / of R. But since / is finitely generated and I Q U/=i Ph we must have / Q Ptl for some n, and thus Pn = Pn + \ = . . . = Q by the minimality of Q over /. If P G Spec(P) and Spec(P) is a tree, then the ascending chain condition on primes implies that P is not the union of primes properly contained in P. This clearly implies that P is minimal over a principal ideal of R. (iii) If P is a prime ideal of R and S is a multiplicative subset of R with S Pi P = 0, then P is minimal over a finitely generated ideal if and only if S~XP is. (3.3) THEOREM. Let fibers B ®R k(P), P G Spec(#), satisfy MFG, then B satisfies MFG. Proof Let Q G Spec(i?) and let P = Q D R. Then P is minimal over some finitely generated ideal I of R since R satisfies MFG. Let S = R — P. Since l l B ®R k(P) = S~ B/S~ (PB) satisfies MFG there exists a finitely generated ideal J Q Q such that (S~lQ)* is minimal over (S-1/)* where the asterisk denotes the image in ] S~ B/S~ \PB). But then Q is minimal over / + /, for if Qx G Spec(£) is such that I + J Q Qx Q Q, then p, = QX n R Q p => px = p. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 353 Thus (S"1/)* Q (^_1Ôi)* £ (S~lQ)* and hence l X S~ Qi = S~ Q^QX = Q. (3.4) COROLLARY. Let R be a ring with the property MFG, and let R^ B be a ring homomorphism. Then B has MFG in the following cases: (a) B is a localization of a finitely generated R-algebra. (b) B is an INC extension of R [3], i.e., no two comparable primes of B contract to the same prime of R. 4. An application to seminormality. In this section we give an application of some of the previous results to an area not directly related to associated primes (or to the MFG condition). Recall that a ring R is said to be reduced if it has no nonzero nilpotents. (4.1) Definition. A reduced ring R is seminormal if every rank 1 projective i£[X|-module is of the form M ®R R[X] for some rank 1 projective i^-module M. This is equivalent to the condition: if a, b e R with a1 = b3 then there exists c (4.2) THEOREM. Let R be a reduced ring with Noetherian spectrum, and let 77 be a finite abelian group of order n. Then RTT is seminormal if and only if R is seminormal, n is regular on R, and nR, nR are radical ideals, where R is the integral closure of R. Proof In [28, Proposition 2.4] one direction has been shown, namely the one assuming that RIT is seminormal. For the converse (see [28, Theorem 2.1] ) the assumption that nR have no embedded primes was needed. By a remark following Proposition 2.4 in [28], this assumption can be removed if we know that weak-Bourbaki primes contract to weak Bourbaki primes under R —» RIT. But this follows from Theorem 2.2 since R satisfies the MFG condition. 5. A result of Northcott. An important application of associated prime ideals is to the study of grade. Let Gr (/; M) denote the polynomial grade of / on M [7]. This is also called the true grade in [22] [23]. The intimate connection of polynomial grade to Ass^ and sK can be inferred from the following well-known result [7] [11]. (5.1) Remark. The following are equivalent for an ideal / and an i^-module M. (i) Gr (/, M) = 0. (ii) Each finitely generated ideal J Q I is contained in a member of Assy(M). Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 354 J. IROZ AND D. E. RUSH (iii) Each finitely generated ideal J Q I is contained in a member of sK(M). The connection between grade and strong Krull primes is further accented by the main result of [23] which states: (5.2) THEOREM. Let A be an R-ideal of positive {true) grade. Then A is projective if and only if (i) A has a resolution of finite length by finitely generated projective modules, and (ii) every P e sK(R/A) has (true) grade one. In this section we show that this theorem remains valid if sK(R/A) is replaced by Assf(R/A). Since it is the if direction of the above theorem that requires the work, then this replacement can be regarded as a strengthening of Northcott's result. Our argument will procède along the same lines as Northcott's. However we can use Theorem 2.6 to reduce to the case that A contains a regular element. Thus one only needs the MacRae invariant for the case considered in [26], [34], or [22], where it was shown that for each torsion 7^-module M having a resolution of finite length by finitely generated projective P-modules, there is a smallest invertible ideal G(M) containing the initial Fitting invariant F(M) of M. Further, given a short exact sequence of such P-modules 0 —» M —» M —> M" —» 0 one has G(M) = G(Mf)G(M"). We will need the following lemma which extends [23, p. 213, Lemma 4] by replacing sK(R/A) by AsSf(R/A) and R by an arbitrary ^-module M. The proof is essentially the same as that given in [23]. (5.3) LEMMA. Let A ¥= R be an ideal and M an R-module. Then Gr (A; M) = inf {Gr (P; M)\P e Assf(R/A) }. Proof. By [21, p. 152, Theorem 16], there exists a prime ideal Q of R such that A Q Q and Gr (A; M) = Gr (g; M). Let P be a minimal prime ideal of A with P Q Q. Then P e Assf(R/A) and Gr (A; M) = Gr (P; M). Note. The above proof actually shows that Gr (v4; M) = inf {Gr (P; M)\P is minimal over a}. (5.4) THEOREM. Let A be an ideal of R with Gr (A; R) > 0. Then A is projective if (and only if) the following conditions hold: (i) Gr (P; R) = 1 for all P G Assf(R/A), Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 355 (ii) A has a resolution of finite length by finitely generated projective R-modules. Proof By adjoining an indeterminate we may assume that there exists a regular element a From [26, Lemma 2.5] we have PdRp(RP/AP) = 1 and thus AP = F(RP/Ap) = G(RP/Ap) = G(R/A)RP [26]. Therefore G(R/A) ¥= R. Now to show that A is projective it suffices to show that ^ = G(R/A). If A ¥= G(R/A) let E = G(R/A)/A. Then £ is a torsion .R-module having a resolution of finite length by finitely generated projective /^-modules, and from the exact sequence 0 -> E -> RIA -> R/G(R/A) -> 0 we get G(#/,4) = G(E)G(R/A). Therefore G(E) = R. By [23, Theorem 8] there exists a minimal prime P of Ann (E) with Gr (P,R) = Gr (P*„, /?„). But then P e Assy(£) c Assf(R/A) and thus we get PdRp(EP) = PdRp{EP)+ Gr (PPP; £P) = Gr^p (PPP; PP) = 1, the second equality coming from [22, Theorem 2, p. 176] and the third by hypothesis. Thus G(EP) = F(EP), [22], [26], and F(EP) = Ann (EP) since EP is cyclic. Thus G(EP) Q PRp, and this contradicts the previous conclusion that G(E) = R since G(E)RP = G(EP) [26]. 6. Associated prime systems. In this section we add some remarks on associated prime systems in general which we hope will give some perspective to the various possible notions of associated prime ideal. For this discussion an axiomatic description of associated primes is conven ient. Let R be a ring and let ^be a mapping which associates to each P-module M a subset s/(M) of Spec(P). (We will write s/R(M) if it is necessary to emphasize the ring R.) Examination of the notions of associated primes that have been used in the past shows the particular significance of the following properties for such a function: (Al) U s/(M) Q ZR(M). (A2) If 0 —> M'—> M —> M" —> 0 is an exact sequence of P-modules, then Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 356 J. IROZ AND D. E. RUSH (A3) If S is a multiplicative subset of i?and?e s/(M) with P n S = 0, then P e s/(M/M(S) ). (A4) If S is a multiplicative subset of i^andPE j/(M/M(S) ), then P G j/(M). (A5) stf{M) = 0 if and only if M = 0. Definition. If the function J/satisfies the above properties we will call s/ an associated prime system for R. A function s# which satisfies the first three of the above properties will be called a formal associated prime system. Axiomatic definitions of associated primes have previously been considered by Merker [18] [19] and McDowell [17]. McDowell's notion of associating system is equivalent to our formal associated prime system. However, we feel that our definition is simpler and more natural since it is given in terms of a single ring P, whereas McDowell's definition requires the simultaneous consideration of all rings of quotients of R. It is well-known (e.g. [2, Chapter IV] ) that Ass is a formal associated prime system and that Assy* is an associated prime system. It is also straight-forward that Ne and Z—S are formal associated prime systems and that sK is an associated prime system. The following result shows that sK and Assy* are the largest and smallest associated prime systems respectively. The first statement is due to McDowell [17]. We repeat the proof in our slightly different setting. (6.1) THEOREM. If stfis a formal associated prime system for R and M is an R-module, then s/{M) Q sK(M). Each member of s/(M) contains a member ofAssf(M) and ifs/is an associated prime system, then Assy(M) Q Proof. Let P e jtf(M). First assume that P = ZR(M). Let / be a finitely generated ideal contained in P. We want to show that there exists a non-zero m e M such that / Q ann (m) Q P. We will proceed by induction on the number of generators n of /. For n = 1 this is clear so assume that it is true for n and let / = (x\, . . . , xn+\), F = (x\, . . . , xn). Let S = {4+1|£>0}. Then JCW + 1 £ Z(M/M(S)) and P £ s/(M/M(S)). By (A2) P <= s/(M(S) ) and P = ZR(M(S) ). By the induction hypothesis there is a non-zero m e M(S) such that /' Q ann (m). Since m e M(S) there exists k such that xn + xm = 0. Choose k minimal, so that xn+]m ^ 0. Then Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 357 k / ç ann (x n + \m) Q P and P G SK(M). For the general case assume that P G S/(M) and let S = R — P. Then P n S = 0 and by (A3) P G s/(M/M(P)) and P = ZR(M/M(P)). By the first part P G sK(M/M(P) ). Therefore P G SK(M) since strong Krull primes satisfy (A4). For the second statement let P G jtf(M). Then since s/(M) Q sK(M) ç Supp (M), then P contains a minimal member Q of Supp (M). But then Q G Assy(M). For the final statement let P G Assy(M), say P is minimal ovet the annihilator of Rm = N, m G M. Then Supp (N/N(P)) = {P} and so from A5 we must have P e s/(N/N(P) ). But then P G J^(A0 Ç J/(M) by A4 and A2. The following methods were suggested by [16] and [19] respectively for obtaining other associated prime systems. (1) If j^is an associated prime system for R and M is an P-module \ets/'(M) consist of those primes of R which are directed unions of elements of s/(M). Is it known that the inclusions Ass/(M) Q st\M) Q sK(M) may be strict where s/ = Assy [17]. (2) Let stf'\M) = {P n R\P G Ass/(M ®R B) for some flat P-algebra B}. It would be interesting to know if jtf" = sK. The next theorem shows how some of the standard results on AsSf extend immediately to arbitrary associated prime systems. (6.2) THEOREM. Let s/be an associated prime system for R and let M be an R-module. Then (i) ZR(M) = U jtf(M) (ii) n s/(M) = {a G R\for each m G M there exists n > 0 such that anm = 0} (= the set of locally nilpotent elements on M). (iii) If S is a multiplicative subset of R and X = {P G s/(M)\P n S = 0}, Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. 358 J. IROZ AND D. E. RUSH then sf(M/M(S)) = X and s/(M(S)) = s/(M) ~ X. Proof, (i) follows since ZR(M) = U Ass/(M) Q u ss?(M) Q ZR(M). To prove (ii) let / be the ideal of locally nilpotent elements on M. Then we have n ^f(M) Q n Ass/(M) = / [2, p. 289]. Further, if a £ p for some P G stf(M) let S = {an\n > 0}. Then PnS = 0 so PG s/(M/M(S) ) by A3. Thus M ± M(S) and therefore a is not locally nilpotent on M. Thus Pi s#(M) = I. To prove (iii) let P G S/(M/M(S) ). Then P G S/(M) by A4. Thus P G X Let P G s/(M(S) ). If P n S = 0 we get the contradiction P G j/(M(S)/M(S')(S'))= J/(0) = 0. Thus PDS ¥> 0 and P G j/(M) - X The result now follows from A2. We conclude this section with a brief discussion of the behavior of associated prime systems under localization. The reader will have noticed the absence of any need to localize thus far in this section since all that was needed was already built in. In fact an associated prime system on R determines in a natural way an associated prime system on each localization of R as follows: Let S be a multiplicative subset of R and let <|>:P —> S~]R be the canonical map. If s?R is a formal associated prime system for R and M is an S~ ^-module define ] S*S-*R(M) = {S P\P G s/R(+M) }. PROPOSITION, (a) Ifs/R is a formal associated prime system for P, the induced map s#s~xR *s a formal associated prime system for S~lR and for each R-module M we have ] 1 -1 ^S- R(S" M) D {S P|P G s?R(M) and P n S = 0}. (b) Ifs/R is an associated prime system for P, thenstfs-\R is an associated prime system for SlR and for each R-module M we have 1 ] [ ^S- R(S~ M) = {S~ P\P <=s/R(M)andP n S = 0}. Proof Let s/R be a formal associated prime system for P. That the induced mapj/5-1^ satisfies Al and A2 is immediate. To show that j/ç-i/j satisfies A3 we will use the following simple result. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 18:55:49, subject to the Cambridge Core terms of use. ASSOCIATED PRIME IDEALS 359 LEMMA. If T is a multiplicative subset of S~ R and T] = {r G R\r/s G Tfor some s G S}, then (ST^-^R = T~\S']R), and for each S~ R-module 4M(T) ) = ^(sr,). Now to show that A3 holds let T be a multiplicative subset of S~ ]R and X l let T, be as in the lemma. If S~ P G S/S -\R{M\ P G Spec(P), with S~ P n T = 0, then P n (STO = 0. Thus P es/ai+MZ+MiSTO) since P G J^R(^M) and J^ satisfies A3. Thus P G stR(+M/£M(T) ] ) = sfR(£M/M{T) ] ) and therefore S~lP G j^-i^(M/M(r)). For the last statement of part (a) let P G S/R(M) with P n 5 = 0. Then by A3 P G j*fc(Af/M(S)) and since M/M(S) c S_1M we get P G ^(S^M)). Thus _1 _1 S P G j/s-i^(5 M). The arguments for part (b) are similar. REFERENCES 1. B. Banaschewki, On the components of ideals in commutative rings, Arch. Math 12 (1961), 22-29. 2. N. Bourbaki, Commutative algebra (Addison Wesley, Reading, Mass., 1972). 3. D. Dobbs, Lying-over pairs of commutative rings. Can. J. Math. 33 (1981), 454-475. 4. P. Dutton, Prime ideals attached to a module, Quart. J. Math. Oxford 29 (1978), 403-413. 5. W. Heinzer, and J. Ohm, Locally Noetherian commutative rings, Trans. Amer. Math. Soc. 755(1971), 273-284. 6. R. Heitmann, Prime divisors and flat extensions, J. of Alg. 74 (1982), 293-301. 7. M. 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